:uhh: A basketball player throws the ball at a 38 degree angle to a hoop which is located a horizontal distance L=7.0m from the point of release, and at a height h=0.3m above it. What is the required speed if the basketball is to reach the hoop?

Use the equation for the projectile (it should be in your text book), then plug in the numbers and solve for the required speed.

cepheid
Staff Emeritus
Gold Member
Strategy...as I mentioned to you in your true/false thread, horizontal and vertical components of the bball's velocity are independent. So...call the initial velocity you're trying to solve for v_0

The x-component of the v_0 is unaffected throughout the journey. Therefore, you don't have a clue yet what to do with it. It is constant, so you need to know the time required for the ball to get there. That depends on how high it needs to go before stopping (the vertical component). So, just note for now that the x component is given by :

$$v_0\cos(38^o)$$

and ignore it for the time being.

Now, the vertical component of the velocity:
$$v_0\sin(38^o)$$

is affected by the gravitational force, which causes a constant downward acceleration, which slows the ball's velocity by 9.81 metres per second, every second. So, how do you solve for the required initial vertical velocity, and the time needed to get the ball up to that height? There are three equations...one gives the vertical component of the velocity as a function of time, (which is useful to you, because you know the final vertical velocity is zero), the other gives the height h(t) as a function of time, in terms of the vertical velocity and and accelaration. The third is the fact that:

$$[v_0\cos(38^o)]t$$ must be equal to three metres.

Hopefully you know which kinematic formulas I'm talking about.

i tried doing this problem by plugging in values into this equation:
Vxo = Vx(cos38) Vyo=Vy(sin38)

combining the two I found,

y=(tan38)x - (g / 2(Vo^2)(cos38)^2)x^2

( I tried your method aswell cepheid - but it was long and complicated - I think I got lost in the numbers along the way.) :s

just wanna make it clear
did you mean
$$y = tan(38)x - \frac{g x^2}{2 v_{0}^2 cos^2(28))$$
if yes, your answer should be fine, what numerical v you got?

sort of yeah - except in the second part it would be (cos38)^2
and whats frac?

I found Vo = 6.115 m/s and it was wrong...:(

that was a typo
here you go
$$y = tan(38)x - \frac{g x^2}{2 v_{0}^2 cos^2(38)}$$

let me check... 1 minute

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how did you write it out so clearly??

i don't get your answer... let me try to figure out what you diid wrong...

yes! thats the equation I used!!

was it just a calculation error then? this is the right equation to use to solve the problem?

I'll try it again - maybe i din't put brackets around certain #'s in my calculator or something...

basically I plugged in:
x = 7.0m
y = 0.3m
g = 9.8m/s^2
and solved for Vo

what is your Vo equal to... you might either have some algebra error or calculatioin error

31.95m/s ??

does that sound more feasible?

i got 5 point something...
sory, can't give you the answer, try again

what am I doing wrong?

~ I only have 2 tries left :S

ok
well at least I know the approximate range of the answer right? hehe

Vo = \sqrt{(tan(38)(7.0m) - \frac{(-9.8m/s^2) (7.0m)^2}{2 (0.3m) cos^2(38)}}

well that didn't work out...

here's how I manipulated the equation:

Vo = Square root of { tan(38)x - ((-g)x^2 / 2ycos(38)^2) }

completely wrong. i believe you can do this problem your own.... just simple algebra.

oh gosh - okay - I'm sure I can - thanks for the help - I'll post my answer if I ever get around the 5 range! lol

g wpuld be a negative number right?

(-9.8m/s^2 as opposed to 9.8m/s^2)

y=(tan38)x - (g / 2(Vo^2)(cos38)^2)x^2
the negative sign has already been taken care of in the equation, g is positive

okay - that makes sense.
I keep getting 8.65m/s